All categories

1 year ago

6. During some actual expansion and compression processes in piston–cylinder devices, the gases have been

1 answer:

5 0

During some actual expansion and compression processes in piston-cylinder devices, the gases have been are the P1= P2.

<h3>What is the pressure?</h3>

Pressure is something that has the pressure that is physical and that causes the pressure is piston-cylinder devices.

During a few real enlargements and compression procedures in piston-cylinder devices, the gases were located to meet the connection PV n = C, wherein n and C are constants.

You might be interested in

Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid–vapor mixture at a pressure

Klio2033 [76]

Answer:

A) ΔS_refrigerant = 0.70754 Kj/K

B) ΔS_space = -0.68441 Kj/K

C) ΔS_total = 0.02313 Kj/K

Explanation:

A) From he table attached, at Pressure of 140 KPa, and by interpolation, we get, Temperature of T = -18.77°C

Converting to degree kelvin yields;

T = -18.77 + 273 = 255.23 K

Formula for entropy change of refrigerant is given as;

ΔS_refrigerant = Q_in/T_refrigerant

We are given Q = 180 KJ

Thus, ΔS_refrigerant = 180/255.23 = 0.70754 Kj/K

B) Formula for entropy change of cooled space is given as;

ΔS_space = Q_out/T_s pace

T_space = -10°C = 273 - 10 = 263K

Thus, ΔS_space = -180/263 = -0.68441 Kj/K

C) the total entropy change would be;

ΔS_total = ΔS_refrigerant + ΔS_space

Thus,

ΔS_total = 0.70754 - 0.68441 = 0.02313 Kj/K

8 0

2 years ago

A factory hires a consultant to recommend ways to improve its productivity. The consultant notices that the production floor is

Romashka [77]

Answer:

Explanation:

Keep only what is used. Remove all unnecessary items. Red tag these items for review.

8 0

2 years ago

A hydraulic jump is induced in an 80 ft wide channel. The water depths on either side of the jump are 1 ft and 10 ft. Please cal

Sphinxa [80]

Answer:

a) The velocity is 42.0833 ft/s

b) The flow rate is 3366.664 ft³/s

c) The Froude number is 0.2345

d) The flow energy dissipated (expressed as percentage of the energy prior to the jump) is 18.225 ft

e) The critical depth is 3.8030 ft

Explanation:

Given data:

80 ft wide channel, L

1 ft and 10 ft water depths, d₁ and d₂

Questions: a) Velocity of the faster moving flow, v = ?

b) The flow rate (discharge), q = ?

c) The Froude number, F = ?

d) The flow energy dissipated, E = ?

e) The critical depth, dc = ?

a) For the velocity:

$\frac{d_{2} }{d_{1} } =\frac{1}{2} (\sqrt{1+8F^{2} } -1)$

$10*2=\sqrt{1+8F^{2} } -1$

Solving for F:

F = 7.4162

$v=F\sqrt{gd_{1} }$

Here, g = gravity = 32.2 ft/s²

$v=7.4162*\sqrt{32.2*1} =42.0833ft/s$

b) The flow rate:

$q=v*L*d_{1} =42.0833*80*1=3366.664ft^{3} /s$

c) The Froude number:

$v_{2} =\frac{q}{L*d_{2} } =\frac{3366.664}{80*10} =4.2083ft/s$

$F=\frac{v_{2}}{\sqrt{gd_{2} } } =\frac{4.2083}{\sqrt{32.2*10} } =0.2345$

d) The flow energy dissipated:

$E=\frac{(d_{2}-d_{1})^{3} }{4d_{1}d_{2}} =\frac{(10-1)^{3} }{4*1*10} =18.225ft$

e) The critical depth:

$d_{c} =(\frac{(\frac{q}{L})^{2} }{g} )^{1/3} =(\frac{(\frac{3366.664}{80})^{2} }{32.2} )^{1/3} =3.8030ft$

4 0

3 years ago

In___cutting, the torch is held and moved by a mechanical device and manually adjusted by the welder.

natita [175]

It is a metal because is saying that is a manually device

6 0

2 years ago

What is the current and power in a 120V circuit if the resistance 5Ω?

podryga [215]

Answer:

24A

Explanation:

trust me

3 0

2 years ago

Read 2 more answers